- #1

- 315

- 0

Prove whethere is linearly independent or not the following:

{[tex]e^x,e^{2x}[/tex]}

{[tex]e^x,e^{2x}[/tex]}

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter evagelos
- Start date

- #1

- 315

- 0

Prove whethere is linearly independent or not the following:

{[tex]e^x,e^{2x}[/tex]}

{[tex]e^x,e^{2x}[/tex]}

- #2

- 3,474

- 257

[tex]a e^x + b e^{2x} = 0[/tex]

for all x. Can you find such scalars?

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

- #4

- 312

- 0

[tex]a e^x + b e^{2x} = 0[/tex] <==> [tex]a e^x = - b e^{2x}[/tex] <==> [tex]ln (a e^x) = - ln (b e^{2x})[/tex] <==> ax = -2bx

So if we carefully choose a, we can definitely show your given set is linearly dependent.

- #5

- 3,474

- 257

I think it's possible:

[tex]a e^x + b e^{2x} = 0[/tex] <==> [tex]a e^x = - b e^{2x}[/tex] <==> [tex]ln (a e^x) = - ln (b e^{2x})[/tex] <==> ax = -2bx

So above, if we carefully choose a, we can definitely show your given set is linearly dependent.

Except

[tex]\ln(a e^x) = \ln(a) + x[/tex]

[tex]\ln(-b e^{2x}) = \ln(-b) + 2x[/tex]

so you need

[tex]\ln(a) - \ln(-b) = x[/tex] for all x!

- #6

- 312

- 0

Except

[tex]\ln(a e^x) = \ln(a) + x[/tex]

[tex]\ln(-b e^{2x}) = \ln(-b) + 2x[/tex]

so you need

[tex]\ln(a) - \ln(-b) = x[/tex] for all x!

Bummer, so sorry.

- #7

- 315

- 0

[tex]a e^x + b e^{2x} = 0[/tex]

for all x. Can you find such scalars?

Put x=0,then a+b=0 ====> a=-b

CORRECT???

THANKS

- #8

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.

- #9

- 315

- 0

No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.

Mat grime ,

- #10

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

Apologies, I missed out an 'in'. The functions e^x and e^2x are clearly linearly independent (over R), which is where I misunderstood what you're saying. (If you don't see it then just rearrange ae^x +be^2x=0 to see that you're claim of linear dependence implies that e^x = -a/b for all x.)

Last edited:

- #11

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.Put x=0,then a+b=0 ====> a=-b

Hence the above are linearly dependent

CORRECT???

THANKS

Another way to prove it is this: if [itex]ae^x+ be^{2x}= 0[/itex] for all x, then it is a constant and its derivative must be 0 for all x: [itex]ae^x+ 2be^{2x}= 0[/itex]. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.

- #12

- 315

- 0

No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.

Another way to prove it is this: if [itex]ae^x+ be^{2x}= 0[/itex] for all x, then it is a constant and its derivative must be 0 for all x: [itex]ae^x+ 2be^{2x}= 0[/itex]. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.

Is not the definition for linear indepedence the following:

for all a,b and x ,real Nos and [tex]ae^x + be^{2x}=0\Longrightarrow a=0=b[/tex]

- #13

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

- #14

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

- #15

- 315

- 0

Is not the definition for linear indepedence the following:

for all a,b and x ,real Nos and [tex]ae^x + be^{2x}=0\Longrightarrow a=0=b[/tex]

can't"prove [itex]e^x[/itex] and [itex]e^{2x}[/itex] are dependent": they areindependent.

But the negation of the above definition is:

HENCE if we put a=2 ,b=-2 ,x=0 ,we satisfy the

- #16

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

the negation of which is

(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

the negation of which is

(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.

Last edited:

- #17

- 315

- 0

That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

.

Can you put that into a quantifier form??

- #18

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

You're making a very easy question seem very hard: can you find real numbers a and b not both equal to 0 so that

ae^x + be^2x

is the zero function? No - it has been proven several times in this thread.

- #19

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

- #20

- 315

- 0

That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

the negation of which is

(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.

The negation of :

(for all a,b)(ae^x + be^2x =0for all x ===.> a=b=0) is

(there exist a,b) [~(ae^x + be^2x = 0

(there exist a,b) [ ae^x = be^2x =0

The negation of [tex]\forall xPx[/tex] is : [tex]\exists x\neg Px[/tex] .

In our case Px is ( ae^x + be^2x =0 ======> a=b=0) and the negation of Px is:

ae^x = be^2x =0 and a=/=0,b=/=0

When you want to show that the function ae^x +be^2 is zero for all xεR ,YOU write :

for all xεR, ae^x + be^2x =0 OR ae^x + be^2x = 0, for all x.

- #21

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

You did not quote all of what he said. He said "The for all x is not part of the definition ofThe negation of :

(for all a,b)(ae^x + be^2x =0for all x ===.> a=b=0) is

(there exist a,b) [~(ae^x + be^2x = 0for all x====> a=b=0)] ======>

(there exist a,b) [ ae^x = be^2x =0there exist xAND a=/=0 ,b=/=0]

Whether you put the quantifier infront ,like i did, or at the end ,like you did we still have the same result.

The negation of [tex]\forall xPx[/tex] is : [tex]\exists x\neg Px[/tex] .

In our case Px is ( ae^x + be^2x =0 ======> a=b=0) and the negation of Px is:

ae^x = be^2x =0 and a=/=0,b=/=0

When you want to show that the function ae^x +be^2 is zero for all xεR ,YOU write :

for all xεR, ae^x + be^2x =0 OR ae^x + be^2x = 0, for all x.

To say that: for all x, is not part of the definition ,then what is part of??

We are talking about functions of x. Saying that "[itex]ae^x+ be^{-x}= 0[/itex]" means that the function [itex]f(x)= ae^x+ be^{-x}[/itex] is equal to the "0 function": g(x)= 0.

- #22

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

- #23

- 2

- 0

- #24

- 315

- 0

You did not quote all of what he said. He said "The for all x is not part of the definition oflinear (in)dependence" and then followed with "it is part of the definition of what it means for the function to be zero" and that was the point of my last response:

We are talking about functions of x. Saying that "[itex]ae^x+ be^{-x}= 0[/itex]" means that the function [itex]f(x)= ae^x+ be^{-x}[/itex] is equal to the "0 function": g(x)= 0.

What is your definition of linear Independence ,in symbolical form or not ??

Last edited:

- #25

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

[tex]\{v_1, v_2, \cdot\cdot\cdot, v_n\}[/tex]

is

If the only set of scalars

[tex]\{a_1, a_2, \cdot\cdot\cdot, a_n\}[/tex]

such that

[tex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0[/tex]

is

[tex]a_1= a_2= \cdot\cdot\cdot= a_n= 0[/itex]

But the point you seem to be having trouble with is this: Since the left side of the first equation is a linear combination of vectors, the "0" on the right side is the 0

When we talk about functions being "independent" or "dependent" we are talking about the functions as members of some vector space- the set of all polynomials, continuous functions, differentiable functions, infinitely differentiable functions, etc. and the 0 vector is the 0

Given almost any set of functions, you can

- #26

- 315

- 0

O.K here is a proof having the

Let b=1

Let a= -e^x

AND [tex]ae^x + be^{2x} = e^x( a + be^x) = e^x( -e^x + e^x) = e^x.0 = 0[/tex]

Hence we have proved there exist [tex]a\neq 0,b\neq 0[/tex] and such that :

[tex] ae^x + be^{2x} = 0[/tex]

- #27

- 315

- 0

: can you find real numbers a and b not both equal to 0 so that

ae^x + be^2x

is the zero function? No - it has been proven several times in this thread.

If you cannot find them it does not mean they do not exist

- #28

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

How do you get that last step? Did you set [itex]a= -e^x[/itex] and b= 1? Of course, you can't do that. [itex]-e^x[/itex] is not aO.K here is a proof having thefor all xpart:

Let b=1

Let a= -e^x

AND [tex]ae^x + be^{2x} = e^x( a + be^x) = e^x( -e^x + e^x)[/tex]

[tex] = e^x.0 = 0[/tex][tex]for all x\in R[/tex]>

Hence we have proved there exist [tex]a\neq 0,b\neq 0[/tex] and such that :

[tex] ae^x + be^{2x} = 0[/tex]for all x

Last edited by a moderator:

- #29

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

You're making a very easy question seem very hard: can you find real numbers a and b not both equal to 0 so that

ae^x + be^2x

is the zero function? No - it has been proven several times in this thread.

I can only conclude that you have not understood anything anyone has said here. matt grime is saying clearly here that it has been proven several time in this thread that it isIf you cannot find them it does not mean they do not exist

- #30

- 315

- 0

I can only conclude that you have not understood anything anyone has said here. matt grime is saying clearly here that it has been proven several time in this thread that it isimpossibleto find such a and b. Your statement here is not at all responsive to that.

Proofs in forums ,Universities ,books, and in general in mathematical literature are

AND to check them whether are correct or wrong

How can a,and b be constants when they can be quantified???

Share:

- Replies
- 4

- Views
- 11K